We begin learning to perform long division on the abacus with a single digit divisor. Just like in the standard algorithm, we will transform division problems into a series of subtraction facts as we move the divisor through the dividend. In the following example we will know we have completed the division problem when the subtraction process has reduced the dividend to zero. As we progress to higher levels, we will perform the long division process where we will have to round our answer to some specific number of significant digits. For now let’s work out a basic abacus division example.
Note: the following examples will place the divisor on the abacus for reference but this is not necessary and is a matter of preference. More advance abacus users will usually not place the divisor on the abacus to save time.
In order to carry out division on the abacus we first need to set up the problem on the abacus. We will follow the usual conventions for setting up our division problem that have proved to be most efficient for calculating the quotient. First we place the dividend, in this case 280, on the abacus by convention to the right side of the abacus giving us room to place the quotient to the left of the dividend. It is then recommended to reserve 3 or 4 rods to the left of the dividend where we will place the quotient answer. Finally it is optional to place the divisor on the abacus. If so, place the divisor to the left of the quotient rods.
Now we are ready to begin our division problem. Just like in multiplication, we will work through the dividend 1 digit at a time using 1 divisor digit at a time. Again we will make full use of our 9×9 multiplication facts to determine each subtraction from the dividend. So we start our problem by noticing that our divisor, 8, is larger than the first digit of the dividend 2. So we can not divide 8 into 2. Therefore we will move 1 more digit right on the dividend and consider the first two digits of the dividend 28. Now we know from our multiplication facts that 8×3 = 24 and 8×4 = 32. So the largest number of times we can take 8 from 28 is 3. Therefore we will place our first quotient digit, 3, on the second rod left from the quotient unit rod.
Here we should make clear how we are maintaining correct place value for the quotient as we work through the dividend. Our dividend has 3 digits so the largest our quotient can be is 3 digits. However, in this problem since we could not divide 8 into the 2, the first dividend digit, we moved 1 more digit to the right on the dividend. Therefore we also have to move 1 digit to the right on the quotient. This is why we placed the first quotient digit 3 on the second rod of the quotient, not the third rod from the quotient unit rod. So we know already our answer will only have 2 digits.
Now we subtract 8×3 = 24 from 28 on the dividend by first subtracting 2 from 2 on the third rod and subtracting 4 from 8 using a 5 pair on the second rod leaving 40 on the dividend.
Next we repeat the entire division process by moving 1 more digit to the right on the dividend. We now consider dividing 8 into 40 (because our dividend pointer is now on the dividend unit rod). We know 8×5 = 40 so we will place 5 on the quotient unit rod.
To complete our division problem, we simply subtract 8×5 = 40 from the 40 on the dividend. Now that we have zeroed out the dividend our problem is complete and the quotient answer is 35.